Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $p = \dfrac{9z + 45}{-2z + 8} \div \dfrac{z^2 + 13z + 40}{4z - 16} $
Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{9z + 45}{-2z + 8} \times \dfrac{4z - 16}{z^2 + 13z + 40} $ First factor the quadratic. $p = \dfrac{9z + 45}{-2z + 8} \times \dfrac{4z - 16}{(z + 5)(z + 8)} $ Then factor out any other terms. $p = \dfrac{9(z + 5)}{-2(z - 4)} \times \dfrac{4(z - 4)}{(z + 5)(z + 8)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 9(z + 5) \times 4(z - 4) } { -2(z - 4) \times (z + 5)(z + 8) } $ $p = \dfrac{ 36(z + 5)(z - 4)}{ -2(z - 4)(z + 5)(z + 8)} $ Notice that $(z - 4)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 36\cancel{(z + 5)}(z - 4)}{ -2(z - 4)\cancel{(z + 5)}(z + 8)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $p = \dfrac{ 36\cancel{(z + 5)}\cancel{(z - 4)}}{ -2\cancel{(z - 4)}\cancel{(z + 5)}(z + 8)} $ We are dividing by $z - 4$ , so $z - 4 \neq 0$ Therefore, $z \neq 4$ $p = \dfrac{36}{-2(z + 8)} $ $p = \dfrac{-18}{z + 8} ; \space z \neq -5 ; \space z \neq 4 $